Can someone help please

Physics

1 answer:

sasho [114]2 years ago

7 0

Ans: The velocity of the bobsled at the bottom of the hill is = 57.7 $\frac{m}{s}$

Explanation: (see image attached)

Hope this helps! Good luck with future work! :)

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200 kW of solar radiation is shining on a 300 m^2 parking lot. What is the insulation on the parking lot?

ale4655 [162]

That's "insolation" ... not "insulation".

'Insolation' is simply the intensity of solar radiation over some area.

If 200 kW of radiation is shining on 300 m² of area, then the insolation is

(200 kW) / (300 m²) = (666 and 2/3) watt/m² .

Note that this is the intensity of the incident radiation. It doesn't say anything
about how much soaks in or how much bounces off.

Wait !
I just looked back at the choices, and realized that I didn't answer the question
at all. I have no idea what "1 sun" means. Forgive me. I have stolen your
points, and I am filled with remorse.

Wait again !
I found it, through literally several seconds of online research.

1 sun = 1 kW/m².

So 2/3 of a kW per m² = 2/3 of 1 sun

That's between 0.5 sun and 1.0 sun.

I feel better now, and plus, I learned something.

7 0

2 years ago

The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a

Vera_Pavlovna [14]

Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

$r=\frac{kze^{2} }{mpV^{2} }$ (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

energy of the particle = 32 MeV = 5.12x10⁻¹²J

At the potential energy is zero, all the energy will be kinetic energy:

$E_{k} =\frac{1}{2} m V^{2}$

Where

m = 4 mp = mass of proton

$5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}$

Replacing in equation 1

$r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2} }{2.56x10^{-12} } =7.1x10^{-15} m$

8 0

2 years ago

A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a

Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

$v_x = \frac{x}{t}$